Tìm x, biết:
a) x + 65 = 105
b) x : 3 = 20
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`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!
\(a>\)\(\left(x+2\right)\) thuộc \(Ư\left(20\right)\)
\(\left(x+1\right)\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)
\(+>x+1=1\)
\(\Rightarrow x=0\)
\(+>x+1=2\)
\(\Rightarrow x=1\)
\(+>x+1=4\)
\(\Rightarrow x=3\)
\(+>x+1=5\)
\(\Rightarrow x=4\)
\(+>x+1=10\)
\(\Rightarrow x=9\)
\(+>x+1=20\)
\(\Rightarrow x=19\)
Vậy \(x\in\left\{0;1;3;4;9;19\right\}\)
\(b>\left(x-2\right)\) là ước của 6
\(\left(x-2\right)\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
\(+>x-2=1\)
\(\Rightarrow x=3\)
\(+>x-2=2\)
\(\Rightarrow x=4\)
\(+>x-2=3\)
\(\Rightarrow x=5\)
\(+>x-2=6\)
\(\Rightarrow x=8\)
Vậy \(x\in\left\{3;4;5;8\right\}\)
\(c>\left(2x+3\right)\) là \(Ư\left(10\right)\)
\(\left(2x+3\right)\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)
\(+>2x+3=1\)
\(\Rightarrow x=-1\)
\(+>2x+3=2\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(+>2x+3=5\)
\(\Rightarrow x=1\)
\(+>2x+3=10\)
\(\Rightarrow x=\dfrac{7}{2}\)
Vậy \(x\in\left\{-1;-\dfrac{1}{2};1;\dfrac{7}{2}\right\}\)
\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)
\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)
hay x=1
b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)
c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)
d: \(\Leftrightarrow-8< x< 3\)
hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)
a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)
\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)
b)TH1: \(x\ge3\)
\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)
TH2: \(-5\le x< 3\)
\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)
TH3: \(x< -5\)
\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)
\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)
\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)
\(2.|3x-1|=\dfrac{4}{5}\)
\(|3x-1|=\dfrac{4}{5}:2\)
\(|3x-1|=\dfrac{2}{5}\)
\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)
\(3x-1=\dfrac{2}{5}\)
\(3x=\dfrac{2}{5}+1\)
\(3x=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}:3\)
\(x=\dfrac{7}{15}\)
\(3x-1=-\dfrac{2}{5}\)
\(3x=-\dfrac{2}{5}+1\)
\(3x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:3\)
\(x=\dfrac{1}{5}\)
a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
=>\(13\sqrt{2x}=28\)
=>căn 2x=28/13
=>2x=784/169
=>x=392/169
b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>2*căn x-5=4
=>căn x-5=2
=>x-5=4
=>x=9
c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=>x-2=0 hoặc x+2=1
=>x=-1 hoặc x=2
a) x + 65 = 105
x = 105 - 65
x = 40
b) x : 3 = 20
x = 20 x 3
x = 60
a) x + 65 = 105 b) x : 3 = 20
x = 105 - 65 (Tìm số hạng) x = 20 x 3 (Tìm số bị chia)
x = 40 x = 60